Rotational Kinematics & Dynamics question from JEE Main 2026 (21 Jan, Shift 1)

Answer: (b)

1. Just after the cut, the rod turns about the end still held; that end has no vertical acceleration, so $a_{cm} = \alpha \dfrac{l}{2}$.
2. Torque about CM: $T\dfrac{l}{2} = \dfrac{ml^2}{12}\alpha = \dfrac{ml^2}{12}\cdot\dfrac{2a_{cm}}{l} \Rightarrow T = \dfrac{m a_{cm}}{3}$
3. Translation: $mg - T = m a_{cm} = 3T \Rightarrow T = \dfrac{mg}{4}$
4. So the answer is (b) $mg/4$.